Domain of $x^y$

Question

What is the exact domain of $f(x,y)=x^y$ ?

Is it $D= ( \mathbb{R^{-}} \times \mathbb{Z}) \cup ( \{0\} \times \mathbb{R-\{0\}}) \cup( \mathbb{R^+} \times \mathbb{R})$ ?

Answer 1

Also $f(-\sqrt 2,\dfrac{1}{3})$ is valid but not included by the domain mentioned up in the question. Let's define $$\forall x\in\Bbb R^{-},p,q\in\Bbb Z , q>0 \qquad , \qquad x^{\frac{p}{q}}=(x^p)^{\frac{1}{q}}\qquad , \qquad \gcd(p,q)=1$$then this power operation means only if $$1)\quad \text{p is even}\\2) \quad \text{both p and q are odd}$$then the correct domain is $$D= ( \mathbb{R_{-}} \times \mathbb{A}) \cup ( \{0\} \times \mathbb{R^+}) \cup( \mathbb{R_+} \times \mathbb{R})$$where$$\Bbb A=\{x\in\Bbb Q|x=\dfrac{p}{q},\quad\text{p(q+1) is even and p and q are coprime}\}$$